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Chapter 6: Problem 63

A telephone survey of 1000 randomly selected US adults found that \(31 \%\) of them say they believe in ghosts. \(^{22}\) Does this provide evidence that more than 1 in 4 US adults believe in ghosts? Clearly show all details of the test.

Short Answer

Expert verified
Yes, there is evidence that more than 1 in 4 US adults believe in ghosts, as the p-value derived from the z-score (which is approximately 0.003) is less than the significance level of 0.05, leading us to reject the null hypothesis.

Step by step solution

01

State the Hypotheses

The null hypothesis, denoted \(H_0\), might be that the population proportion \(p\) is equal to 0.25. The alternative hypothesis, denoted \(H_a\), would then be that the population proportion \(p\) is greater than 0.25. In symbols: \(H_0: p = 0.25\) and \(H_a: p > 0.25\).
02

Calculate the Test Statistic

The test statistic is a z-score (z). The formula for calculating z is: \(z = ( \hat{P} - P_0 ) / \sqrt{ (P_0 * (1 - P_0)) / n }\) , where \(\hat{P}\) is the sample proportion, \(P_0\) is the population proportion under the null hypothesis, and \(n\) is the sample size. Substituting values, we get: \(z = (0.31 - 0.25) / \sqrt{ (0.25 * (1 - 0.25)) / 1000 } = 2.76.\)
03

Determine the Critical Value and Make a Decision

For a one-tailed test at a significance level of 0.05, the critical value for z is 1.645. Our calculated z-value of 2.76 falls into the rejection region (z > 1.645), therefore, we reject the null hypothesis. Which means, we have enough evidence at the 5% significance level to say that more than 1 in 4 US adults believe in ghosts.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

z-score
The z-score is a key concept in hypothesis testing. It measures how many standard deviations a data point (or statistic) is from the mean under the null hypothesis. In our example, we're trying to determine if the sample proportion of US adults who believe in ghosts is significantly different from 25%.

To calculate the z-score, we use the formula:
  • \(z = ( \hat{P} - P_0 ) / \sqrt{ (P_0 * (1 - P_0)) / n }\)
Where:
  • \(\hat{P}\) is the sample proportion (0.31)
  • \(P_0\) is the hypothesized population proportion (0.25)
  • \(n\) is the sample size (1000)
Applying our values, we find the z-score to be 2.76. This result tells us that the sample proportion is 2.76 standard deviations above the mean proportion under the null hypothesis. It's crucial because it determines whether we have sufficient evidence to reject the null hypothesis.
population proportion
When discussing hypothesis testing, the population proportion is a fundamental concept. It's the percentage of individuals in a population who have a particular attribute. In this scenario, the population proportion pertains to the percentage of US adults who believe in ghosts.

In hypothesis testing, we often set up a model where we propose a specific value for the population proportion known as the null hypothesis. Here, this value was 0.25, suggesting that we think at least 25% of the population believes in ghosts. However, our sample found 31%, which we represent mathematically as the sample proportion \(\hat{P}\).

The goal is to use the sample data to determine if the observed proportion (\(\hat{P}=0.31\)) provides strong enough evidence to conclude that the true population proportion differs from the hypothesized 0.25. It's all about measuring if our sample is enough to reflect a different reality for the large population.
significance level
The significance level is a critical component of hypothesis testing that represents the probability of rejecting the null hypothesis when it is actually true. It's denoted by \(\alpha\) and is typically chosen by the researcher before conducting the test.

In many contexts, a common significance level is 0.05, meaning there's a 5% chance of committing a Type I error, which is rejecting a true null hypothesis. In this example, the significance level is indeed set at 0.05. This is interpreted as being willing to take a 5% risk that we have incorrectly claimed that more than 25% of US adults believe in ghosts, based on our sample analysis.

The chosen significance level also determines the critical value or threshold for the z-score in our case. For a one-tailed test at 0.05 significance, the critical z-value is 1.645. Since our calculated z-score of 2.76 exceeds this critical value, we conclude that we have enough evidence to reject the null hypothesis.

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Most popular questions from this chapter

We see in the AllCountries dataset that the percent of the population that is elderly (over 65 years old) is 17.0 in Austria and 15.9 in Denmark. Suppose we take random samples of size 200 from each of these countries and compute the difference in sample proportions \(\hat{p}_{A}-\hat{p}_{D}\), where \(\hat{p}_{A}\) represents the proportion of the sample that is elderly in Austria and \(\hat{p}_{D}\) represents the proportion of the sample that is elderly in Denmark. Find the mean and standard deviation of the differences in sample proportions.

Data 1.3 on page 10 discusses a study designed to test whether applying a metal tag is detrimental to a penguin, as opposed to applying an electronic tag. One variable examined is the date penguins arrive at the breeding site, with later arrivals hurting breeding success. Arrival date is measured as the number of days after November 1st. Mean arrival date for the 167 times metal- tagged penguins arrived was December 7 th ( 37 days after November 1 st ) with a standard deviation of 38.77 days, while mean arrival date for the 189 times electronic-tagged penguins arrived at the breeding site was November 21 st (21 days after November 1 st ) with a standard deviation of \(27.50 .\) Do these data provide evidence that metal-tagged penguins have a later mean arrival time? Show all details of the test.

Involve scores from the high school graduating class of 2010 on the SAT (Scholastic Aptitude Test). The distribution of sample means \(\bar{x}_{m}-\bar{x}_{f},\) where \(\bar{x}_{m}\) represents the mean Critical Reading score for a sample of 50 males and \(\bar{x}_{f}\) represents the mean Critical Reading score for a sample of 50 females, is centered at 5 with a standard deviation of \(22.5 .\) Give notation and define the quantity we are estimating with these sample differences. In the population of all students taking the test, who scored higher on average, males or females?

Refer to a study on hormone replacement therapy. Until 2002 , hormone replacement therapy (HRT), taking hormones to replace those the body no longer makes after menopause, was commonly prescribed to post-menopausal women. However, in 2002 the results of a large clinical trial \(^{56}\) were published, causing most doctors to stop prescribing it and most women to stop using it, impacting the health of millions of women around the world. In the experiment, 8506 women were randomized to take HRT and 8102 were randomized to take a placebo. Table 6.16 shows the observed counts for several conditions over the five years of the study. (Note: The planned duration was 8.5 years. If Exercises 6.205 through 6.208 are done correctly, you will notice that several of the p-values are just below \(0.05 .\) The study was terminated as soon as HRT was shown to significantly increase risk (using a significance level of \(\alpha=0.05)\), because at that point it was unethical to continue forcing women to take HRT). Does HRT influence the chance of a woman getting invasive breast cancer? $$ \begin{array}{lcc} \hline \text { Condition } & \text { HRT Group } & \text { Placebo Group } \\ \hline \text { Cardiovascular Disease } & 164 & 122 \\ \text { Invasive Breast Cancer } & 166 & 124 \\ \text { Cancer (all) } & 502 & 458 \\ \text { Fractures } & 650 & 788 \\ \hline \end{array} $$

If random samples of the given sizes are drawn from populations with the given proportions: (a) Find the mean and standard error of the distribution of differences in sample proportions, \(\hat{p}_{A}-\hat{p}_{B}\) (b) If the sample sizes are large enough for the Central Limit Theorem to apply, draw a curve showing the shape of the sampling distribution. Include at least three values on the horizontal axis. Samples of size 80 from population \(A\) with proportion 0.40 and samples of size 60 from population \(B\) with proportion 0.10
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